Wizard X's HI/E Reduction Errors

Elementary · Tue Aug 02, 2005 3:18 pm

After studying Wizard X's write-up in the Rhodium archive I have found some errors, (please correct me if I'm wrong)

Firstly

You can not gain a 57% concentration of hydriodic acid just by the addition of H2S to Iodine in water, you need to distill the solution afterwards.

From General and Inorganic Chemistry - J. R. Partington

A less concentrated solution is made by passing hydrogen sulphide into a saturated solution of iodine and adding more iodine as the reaction proceeds. The reaction stops when the s.g. of the solution reaches 1.56. The solution is filtered and distilled. Very weak acid (with a little hydrogen sulphide) first distills, then the temperature rises sharply to 126C and the 57% acid is collected.

Hydriodic Acid S.G. @ 13c : HI %
1.413 : 40.4, 1.528 : 48.4, 1.603 : 52.4, 1.696 : 57.3

Secondly

(PROCEDURE 2)
150 mls of glacial acetic acid, 15 grams of red phosphorus, and 33.50 grams of iodine.(NOTE 6) The mixture is allowed to react for 15 - 20 minutes, until all of the iodine has reacted. Then add 10 mls of distilled water and mix

The above seems correct, the acetic acid is used as a solvent and only 10ml of water is used. But below (NOTE6) seems incorrect and will not give you anywhere near the HI 55-57% concentration needed.

(NOTE6)
Alternatively, 150 mls of hot distilled water, 15 grams (0.48 moles) of red phosphorus, and 33.50 grams (0.132 moles) of iodine in small 3-5 gram portions is added. The mixture is allowed to react for 15-20 min, until all of the iodine has reacted.

Quite a difference between this and the 10ml of water to 10g of Iodine used in the "begineers long reflux" !

Elementary · Tue Aug 02, 2005 9:04 pm

Properties of 57% Hydriodic Acid

Density Solution = 57%
Specific Gravity = 1.70
Moles/Litre = 7.57
Grams/Litre = 969

http://www.usbweb.com/reference2.asp?id_ref=22

So this tells us that 969g of HI in 1L of water gives a 57% solution of hydriodic acid or 96.9g of HI in 100mL of water.

Atomic weights
H = 1.00797
I = 126.9044
O = 15.9994

1 mole of HI = 127.91237
1 mole of H2O = 18.01534

So if there are 7.57 moles of HI in 1L of water

127.91237 x 7.57 = 968.29664 (969)

As a rule of thumb it is know that 1L of water weighs about 1000g

So how many moles of water in one 1kG ?
1000/18.01534 = 55.50825

So 55.5 moles of water added to 7.6 moles of HI gives us 57% Hydriodic Acid a ratio of 7.3:1

Now lets work out the water to iodine ratio to make 100% HI

2P + 10I + 8H2O = 10HI + 2H3PO4

8 moles of Water = 144.122g
10 moles of Iodine = 1269.044g
10 moles of HI = 1279.1237g

A ratio of 1:0.88 for iodine:water

So 0.88g moles of water (18g) with 1 mole of Iodine (127g) will give us HI 100%

So now we need to add things up

127g of Iodine > 15.85g of Water > 128g of HI
128g of HI > 131.5g of water > 57% Hydriodic Acid

15.85+131.5=147.35 > 147.35g of Water > approx 147.35 ml

End result 127g of iodine to 147.35ml of water (plus the required amount of RP) gives approx 57% Hydriodic Acid.

End ratio, Iodine:Water(g:g or g:ml) > 1:1.16

This my reasoning for why Wizard X's calculations are wrong for procedure 2 note 6.

If you use 33.5 grams of Iodine then 38.86ml of water is needed to achieve a 57% Hydriodic acid solution rather than 150ml as stated.

Ionium · Wed Aug 03, 2005 4:37 am

You're not mistaken about the wiz's HI being nowhere near 57%, 18% or so is more like it. Your calculations also look good(At least I didn't find anything wrong with them, but that's of course no guarantee that no one else will), but you seem to misunderstand or misinterpret the information you found. Let's have a look at that:

Elementary · Wed Aug 03, 2005 4:14 pm

Hmmmm "my brain hurts" back to the books.

I think it would be easier just to tritrate the solution to prove the point, I'm getting lost in calculations.

java · Consumer

Here is a copy of Wizard's procedure on Rhodium's Archives,............java

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Ephedrine Reduction with HI/P
by Wizard X

57% HI Solution (Hydriodic Acid):

Boiling point 125.5-126.5 deg C

Density 1.70 gr/ml

55-57% w/w HI is 0.936 to 0.99 grams of HI per ml.

Reaction: H2S + I2 ===>> 2HI + S

In a fume cupboard a 1.5 three neck flask is added a mixture of 480 grams of iodine and 600 mls of distilled water. The centre neck is added a sealed stirring unit which leads almost to the bottom of the flask. The left neck is fitted and sealed with a glass tube which extends almost to the bottom of the flask but does not touch the stirrer. This is connected to the hydrogen sulphide (H2S) generator. The right neck is fitted and sealed with a short glass tube connected with a plastic tube to the bottom of an inverted funnel in a 5% sodium hydroxide solution.

The mixture is vigorously stirred and a stream of hydrogen sulphide (H2S) gas is passed rapidly in the iodine/water mixture as rapidly as it reacts with the iodine. After several hours, 2-3 hours, the iodine disappears and the liquid assumes a yellow colour (sometimes almost colourless) and most of the precipitated sulphur sticks together in the form of a hard lump.[NOTE 1] This liquid contains a mixture of HI, H2S and S. The liquid is now filtered through a large funnel plugged with glass wool to remove the sulphur S. No need removing the dissolved H2S as this enhances the reductive power. Add a few crystals of iodine to this HI/H2S solution and store at 0-5 deg C.

[NOTE 1] The sulphur lump in the flask can be removed by refluxing with concentrated nitric acid.

Reduction of Ephedrine HCl to methedrine

Ephedrine HCl, molecular weight MW = 201.73 grams/mole

Ephedrine Freebase, molecular weight MW = 165.23 grams/mole

PROCEDURE 1:

A mixture of 80.7 grams (0.4 mole) of EPHEDRINE HCl, 20 grams of red phosphorus and 170 mls of 57% hydriodic acid is refluxed for 25 hours. Stopping refluxing after 25 hours, and allowing to stand at room temperature for another 12 hours. A total of 37 hours. After this time a large amount of crystalline material has seperated out from the solution.

The reaction mixture was diluted with 700 mls of distilled water and filtered to remove the red phosphorus.[NOTE 2] The yellow filtrate was treated with a few crystals of sodium thiosulphate to remove any free iodine.[NOTE 3] It is then made basic with 40 % sodium hydroxide solution.

The liberated freebase methedrine which seperates is solvent extrated with two, 75-100 ml ether portions and the ether/amine solution is first washed with 50 mls of distilled water and the ether/amine solution dried with anhydrous sodium carbonate.[NOTE 4] After removal of the ether, the oil was vacuum distilled at a vacuum of 15 mmHg at 93 degC.[NOTE 5] The yield is 80 - 88%

[NOTE 2] Rinse the red phosphorus with a little distilled water and collect with other filtrate. Use suction filtration.

[NOTE 3] Add one or two crystals at a time and mix. Then add more sodium thiosulphate in one or two crystal portions until the filtrate shows no red or purple iodine.

[NOTE 4] Anhydrous magnesium sulphate can be used and rinse with a little ether to extract all the freebase out.

[NOTE 5] The oil is a mixture of methedrine and iodo-ephedrine. It is very difficult to vacuum distil the methedrine out alone. A better method is to reduce the iodo-ephedrine/methedrine oil residue with lithium aluminium hydride in ether. The iodo-ephedrine is easy reduced further to methedrine.

PROCEDURE 2

In a 500 mls round-bottom flask is added 150 mls of glacial acetic acid, 15 grams of red phosphorus, and 33.50 grams of iodine.[NOTE 6] The mixture is allowed to react for 15 - 20 minutes, until all of the iodine has reacted. Then add 10 mls of distilled water and mix, then add 8.07 grams (0.04 mole) of EPHEDRINE HCl, and mix. A reflux condenser is added to the 500 mls round-bottom flask and the mixture is gently refluxed (gently boiled continuously) for 2 and 1/2 hours. The hot mixture is suction filtered while still hot to remove the excess red phosphorus.[NOTE 7] The hot filtrate is added slowly by pouring into a solution of 20 grams of sodium bisulphite in 1 liter of distilled water. The solution is made basic and solvent extracted as above using two 100 ml portions of ether, washed and dry.[NOTE 4] The oily residue extracted (after removing the ether) is a mixture of methedrine, iodo-ephedrine and minor amount of acetic ester ephedrine.[NOTE 5]

[NOTE 6] Alternatively, 150 mls of hot distilled water, 15 grams (0.48 moles) of red phosphorus, and 33.50 grams (0.132 moles) of iodine in small 3-5 gram portions is added. The mixture is allowed to react for 15-20 min, until all of the iodine has reacted. Then add 8.07 g (0.04 mole) of EPHEDRINE HCl and follow Procedure 1. The yield is 80-88%

[NOTE 7] Have some hot glacial acetic acid on hand to rinse the flask and the filtered red phosphorus.

Calculation of HI to Ephedrine Ratio

In Procedure 1 and 2, two moles of HI react with one mole of Ephedrine giving Methamphamine and I2.

Ephedrine + 2HI ==>> methedrine + I2 + H2O

Therefore the Ephedrine:HI ratio is theoretically 1:2, but to assure that the reaction goes to completion, the actual ratios used above are 1:3.3

This ratios has been calculated as follows:

Procedure 1:

1 ml of 57% HI = 0.99 grams of HI

170 mls of 57% HI = (170 x 0.99) = 168.3 grams of HI = 1.32 moles HI

Since we use 0.4 moles of Eph then (1.32/0.4) = 3.3

Since the reaction of phosphorus, iodine and water is 2 P + 3 I2 + 6 H2O --> 6 HI + 2 H3PO3 the ratio of HI:I2 is 3:3

Procedure 2

For every 0.04 moles of Eph, then (0.04 x 3.3) = 0.132 moles of HI, since the ratio of HI:I2 is 3 : 3 = 1, then we need 0.132 moles of I2 = 33.50 grams of I2.

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java · Consumer

Here is WizardX response to the purported error............java

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http://216.117.165.51/maxwebportal/forum/topic.asp?TOPIC_ID=208&FORUM_ID=5&CAT_ID=1&Topic_Title=Synthetikal+questions+H2S+%2B+I2&Forum_Title=temp+chem+101

or in print if not a traveler............

Java sent me this PM: There is some discussion about one of your writings claiming that you are wrong, I just feel you should know so you can clear it up. https://synthetikal.com/synthforum/viewpost.php?p=8813

quote:
H2S + I2 ===>> 2HI + S
In a fume cupboard a 1.5 three neck flask is added a mixture of 480 grams of iodine and 600 mls of distilled water. The centre neck is added a sealed stirring unit which leads almost to the bottom of the flask. The left neck is fitted and sealed with a glass tube which extends almost to the bottom of the flask but does not touch the stirrer. This is connected to the hydrogen sulphide (H2S) generator. The right neck is fitted and sealed with a short glass tube connected with a plastic tube to the bottom of an inverted funnel in a 5% sodium hydroxide solution.

The mixture is vigorously stirred and a stream of hydrogen sulphide (H2S) gas is passed rapidly in the iodine/water mixture as rapidly as it reacts with the iodine. After several hours, 2-3 hours, the iodine disappears and the liquid assumes a yellow colour (sometimes almost colourless) and most of the precipitated sulphur sticks together in the form of a hard lump.[NOTE 1]
This liquid contains a mixture of HI, H2S and S. The liquid is now filtered through a large funnel plugged with glass wool to remove the sulphur S. No need removing the dissolved H2S as this enhances the reductive power. Add a few crystals of iodine to this HI/H2S solution and store at 0-5 deg C.

[NOTE 1] The sulphur lump in the flask can be removed by refluxing with concentrated nitric acid.

H2S + I2 ===>> 2HI + S

480 of I2
Grams of 1 mole of I2 = 253.8 g/mol

Moles of I2 = 480/253.8 = 1.89 moles

Ratio of I2:HI = 1:2

Therefore the moles of HI is 1.89 x (2/1) = 3.78 of HI

57% w/w HI has a Molarity of 7.58 mol/Lt. That is 7.58 moles of HI in 1 liter of water.
47% w/w HI has a Molarity of 5.5 mol/Lt. That is 5.5 moles of HI in 1 liter of water.

Molarity = moles/Lt

Molarity of HI = 3.78/(600/1000) = 3.78/0.6 = 6.3 mol/Lt.

The procedure above uses 480 grams of iodine and 600 mls of distilled water. This gives a 6.3 mol/Lt solution of HI.
Less than 57% and more 47%.

The concentration of HI is NOT CRITICAL. What is critical is the recycling reagent (eg. P or H3PO2) needed to convert the I2 back to HI during the refluxing.

If you want a 57% HI solution you must distill at the azeotropic temperature of 127 oC.

Download HICalc.zip at http://wizardx.t35.com/archives/

2 P + 3 I2 + 6 H2O reactants can be cancelled down to P + 3/2 I2 + 3 H2O reactants.

Change the 1 to 0.044 in the Mole Factor of equation 2 P + 3 I2 + 6 H2O, then notice HI, it's 0.264 mole and the I2 is 0.132 moles.

Molarity = moles/Lt

Molarity of HI = 0.264/(150/1000) = 0.264/0.15 = 1.76 mol/Lt of HI. The large excess of RP ensures that any I2 liberated is quickly converted back to HI during reflux. Use 0.587 moles of ephedrine with 1.76 mol/Lt of HI. A ratio of 1:3.

Albert Einstein: Great ideas often receive violent opposition from mediocre minds.

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Neo (Red Pill)

Junior Member

100 Posts

Posted - 08/20/2005 : 06:37:54 AM
Elementary's a little late to class.. this has already been flogged to death on the hive.
This method produces strong HI.. good stuff.. but to get it fuming you need to distill it (not very fun) or toss in some RP and I2 into your reflux reaction to push the equilibrium.

An interesting tid bit from the original thread dealt with filtering the sulfur, here WizardX has correctly recommended filtering thru glass wool.

Another version involved the elemental sulfur forming a "puck" that could be plucked from the solution or the acid simply decanted.. which seems like a stretch.

Even more far fetched versions involved filtering the strong acid through coffee filters o.O

H2S is prepared via the acid hydrolysis of ferrous sulphide (FeS). This precursor is easily produced by mixing equal parts iron wool [Fe(II)] and elemental sulfur. This mixture is submersed in water which is then boiled off. Caution: small portions of H2S gas is released during the thermal decomposition of FeS. SWIM should be careful were he stores FeS as it slowly decompose to H2S at room temp.

FeS is then added to a Kipps apparatus which is used to control acid exposure of the sulphretted iron, hopefully preventing a run away reaction.

FeS (s) + 2 HCl (aq) = FeCl2 (aq) + H2S (g)